Question

The position of a particle moving along the x-axis is given by s(t) = 3t^2 + 6.


Use the difference quotient to find the velocity, v(t)=____, and acceleration, a(t)=______.

How do I solve this? PLease explain! Thank you:)

Answer 1

The velocity function is the derivative of position.

We take our difference quotient (finding the rate of change of position over time)\[\Large\bf\sf \frac{s(t+\Delta t)-s(t)}{(t+\Delta t)-t}\quad=\quad \frac{s(t+\Delta t)-s(t)}{\Delta t}\]This gives us the average rate of change of position (velocity) over some time period.

To find velocity as a function, we want the `instantaneous` rate of change.
So we'll let our Delta t approach zero using limits.\[\Large\bf\sf \lim_{\Delta t\to 0} \frac{s(t+\Delta t)-s(t)}{\Delta t}\]This will give us our velocity function.

Answer 2

So ummmm

Answer 3

\[\Large\bf\sf s(t)=3t^2+6, \qquad\qquad s(t+\Delta t)=3(t+\Delta t)^2+6\]
We just need to plug some stuff in.

Answer 4

ohh so how do we figure out what to plug in? :/

Answer 5

Try to get comfortable with the difference quotient :O
It's the same formula you used for slope in your basic algebra class except now we're using function notation so it looks a little different.
\[\Large\bf\sf \color{royalblue}{s(t)=3t^2+6}, \qquad\qquad \color{orangered}{s(t+\Delta t)=3(t+\Delta t)^2+6}\]
We want to plug those two things into our difference quotient below,
\[\Large\bf\sf \lim_{\Delta t\to 0} \frac{\color{orangered}{s(t+\Delta t)}-\color{royalblue}{s(t)}}{\Delta t}\]

Answer 6

\[\Large\bf\sf \lim_{\Delta t\to 0} \frac{\color{orangered}{3(t+\Delta t)^2+6}-\color{royalblue}{(3t^2+6)}}{\Delta t}\]Too much maf? :O
Confusing?

Answer 7

Then from there we simplify.

Answer 8

What are you thinking food? :D

Answer 9

haha ohhh is that the same thing as lim h->0 f(a+h) - f(a) / h right?

Answer 10

yes, the limit definition of the derivative.

Answer 11

erm would 3(t+∆t)(t+∆t)+6-(3t^2+6)/ ∆t

?

Answer 12

and we need to do something to get rid of the ∆t in the denominator right?

Answer 13

We start by `trying to` plug Deltat=0 directly in.
We see that we have a problem, so yes we'll need to do some algebra:
Expand out the brackets, cancel some stuff out, then try to plug Deltat=0 in again.

Answer 14

ohh okay so

3(t+t) +6-3t^2 +6 / 0 ?

Answer 15

Yah, we're dividing by 0, bad bad bad.
Since that didn't work, we have to back up, and simplify the expression before we can plug the zero in.

Answer 16

ohh okay...

would this be the simplified version?

3(t+∆t)(t+∆t)+6-(3t^2+6)
----------------------
∆t

Answer 17

oh wait...

3(t+∆t)(t+∆t)+6-3(t^2+2)
------------------------
∆t

??

Answer 18

You need to multiply out those brackets in the first term.

Answer 19

Don't factor stuff out, try to expand everything.

Answer 20

\[\Large\bf\sf (a+b)^2\quad=\quad a^2+2ab+b^2\]\[\Large\bf\sf (t+\Delta t)^2\quad=\quad ?\]

Answer 21

oh so t^2 + 2t∆t + ∆t^2 ?

Answer 22

\[\Large\bf\sf \lim_{\Delta t\to 0} \frac{{3(t^2+2t\Delta t+\Delta t^2)+6}-(3t^2+6)}{\Delta t}\]Ok good.

Answer 23

what happens from here?

Answer 24

Expaaaand! :O

Answer 25

3t^2-6t∆t+3∆t^2 + 6 - 3t^2 - 6
------------------------------
∆t

?? 3t^2 and 6 cancel out and then you get this?

-6t∆t + 3∆t^2
------------
∆t

??

Answer 26

Why is that second term -6t∆t?
Shouldn't that be +?

Answer 27

oh sorry yes, +

6t∆t + 3∆t^2
------------
∆t
? better? :)

Answer 28

Ok looks good!
Yes good, you remember to distribute your negative sign.

Answer 29

Hmm what next? :O
Any other cancellations we can make?

Answer 30

yay!:)

um

6t+3∆t ?

Answer 31

Ok good! We've simplified it down to this,
\[\Large\bf\sf \lim_{\Delta t\to0}6t+3 \Delta t\]

Answer 32

so 6t + 3(0) ?

Answer 33

Yes, good job, we can try plugging in Deltat=0 now since we've cancelled out some stuff.
No longer giving us a problem.

Answer 34

So we've found that, given our position function:\[\Large\bf\sf s(t)=3t^2+6\]Our velocity function will be:\[\Large\bf\sf v(t)=6t\]

Answer 35

ohh cool!! :) yay!! so now we have to find a(t) right?

Answer 36

Yes. a(t) will follow the same process.

The acceleration function is the derivative of the velocity function,\[\Large\bf\sf a(t)\quad=\quad \lim_{\Delta t\to 0}\frac{v(t+\Delta t)-v(t)}{\Delta t}\]

Answer 37

okay so

6t^2 + 6t∆t - 6t
---------------
∆t

??

Answer 38

Hmm where is the square coming from :o

Answer 39

6t * t ?

Answer 40

\[\Large\bf\sf v(\color{royalblue}{t})=6(\color{royalblue}{t}), \qquad\qquad v(\color{royalblue}{t+\Delta t})=6(\color{royalblue}{t+\Delta t})\]

Answer 41

We plug those in.
No square this time.

Answer 42

ohhh i see now :/ oops!!

so

6(t+∆t) - 6(t)
-----------
∆t

?

Answer 43

ya

Answer 44

okay and that becomes

6t + 6∆t - 6t / ∆t
= 6∆t/∆t = 6 ?

Answer 45

yay good job.\[\Large\bf\sf a(t)=6\]

Answer 46

yay!! thank you!!! :D